Question

$A\, 20$ litre container at $400\, K$ contains $CO _{2}( g )$ at pressure $0.4\, atm$ and an excess of SrO (neglect the volume of solid SrO). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of $CO _{2}$ attains its maximum value, will be (Given that $: SrCO _{3}( s ) \rightleftharpoons SrO ( s )+ CO _{2}( g )$ $\left. K _{ p }=1.6\, atm \right)$

• A. 10 litre
• B. 4 litre
• C. 2 litre
• D. 5 litre

Answer 1

The Correct Option is D

To solve this question, we need to find the maximum volume of the container when the pressure of $CO_2$ gas reaches its maximum value given the equilibrium condition with the reaction:

SrCO_3(s) \rightleftharpoons SrO(s) + CO_2(g)

where the equilibrium constant $K_p = 1.6\, \text{atm}$.

1. Initially, the $CO_2$ pressure in the container is $0.4\, \text{atm}$.
2. According to the equilibrium condition, we know that the maximum pressure of $CO_2$ should equal $K_p$ when equilibrium is reached, which is $1.6 \, \text{atm}$.
3. Using the ideal gas law, if the pressure changes from $0.4 \, \text{atm}$ to $1.6 \, \text{atm}$ at constant temperature, the volume should decrease to maintain equilibrium.
4. The ideal gas law can be used here:
   P_1V_1 = P_2V_2
   where:
   • P_1 = 0.4 \, \text{atm}
   • V_1 = 20 \, \text{litre}
   • P_2 = 1.6 \, \text{atm}
   • V_2 is the new volume we need to calculate.
5. Substituting the known values, we have:
   0.4 \times 20 = 1.6 \times V_2
   V_2 = \frac{0.4 \times 20}{1.6} = 5 \, \text{litre}
6. Therefore, the maximum volume of the container when the pressure of $CO_2$ attains its maximum equilibrium value is 5 \, \text{litre}.

Hence, the correct answer is 5 litre.